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I'm trying to prove:

If $f:A\subset \mathbb{R}^m \to \mathbb{R}^n$ and $A$ is open, then the following statements are equivalent:

1) $f$ is continuous on $A$.

2) $f^{-1}(V)$ is open in $\mathbb{R}^m$, for all $V\subset \mathbb{R}^n$ open.

Thanks for your help.

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It's hard to know how to help in this situation without giving the whole game away. Could you say in a few words what you've tried? –  Dylan Moreland Sep 5 '11 at 0:49
    
@gary That's a good point. I was assuming that "continuous" means "satisfies the $\varepsilon$-$\delta$ condition at every point of $\mathbf R^m$", but it would be nice to spell that out in the question. –  Dylan Moreland Sep 5 '11 at 0:52
    
@Hiperion: As this is tagged "general topology", am I correct in guessing that your definition of "continuous function" is "the inverse image of an open set is open"? –  Arturo Magidin Sep 5 '11 at 1:01
    
My connection is slow today. Please be patient. Thanks –  Hiperion Sep 5 '11 at 1:42
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What is the general definition of continuity you are using? –  gary Sep 5 '11 at 2:49
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2 Answers 2

up vote 1 down vote accepted

Assuming your using the definition of continuity that the inverse image of an open set is open, one way of doing the proof is:

i) Show that if the restriction of $f\colon \mathbb R^m \rightarrow \mathbb R^n:$ to any open subset of the domain is continuous, then f is globally-continuous, meaning f satisfies condition #2. For this, you can (Let X be the domain, mapping into Y to simplify; result generalizes to any two topological spaces X,Y anyway.) consider an open subset W of Y, and consider $f^{-1}(W)$ . By def. , f is continuous if $f^{-1}(W)$ is open in the subspace A of X .

2) If we use 2 for the definition of continuity, assume $f:X \rightarrow Y$ is continuous, so that $f^{-1}(W):=V$ is open in X. Can you show V is open in the subspace A?

3) Can you find an outlet, my battery is dy......

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Thank you very much. –  Hiperion Sep 5 '11 at 1:51
    
No problem, Hiperion. –  gary Sep 5 '11 at 2:47
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This is my answer:

1) $\to$ 2).

If $f^{-1}(V)=\emptyset$ the answer is trivial.

Let $a\in f^{-1}(V)$. Particularly $a\in A$. As $f$ continuous in $A$ then $\forall \varepsilon>0 \exists\delta>0$ such that, if $||x-a||<\delta \implies ||f(x)-f(a)||<\varepsilon$. i.e:

if $x\in \mathbb{B}_{\delta}(a) \implies f(x)\in \mathbb{B}_{\varepsilon}(f(a))$

$\implies f(\mathbb{B}_{\delta}(a))\subset \mathbb{B}_{\varepsilon}(f(a))$

$\implies \mathbb{B}_{\delta}(a)\subset f^{-1}(\mathbb{B}_{\varepsilon}(f(a)))$

$\therefore f^{-1}(V)$ is open.

2) $\to$ 1).

Let $\varepsilon>0$ and $a\in A$. $V$ is open then exists $\mathbb{B}_{\varepsilon}(f(a)))$ open. Now $a\in f^{-1}(\mathbb{B}_{\varepsilon}(f(a))))$ is open. i.e:

$\exists \delta$ such that $\mathbb{B}_{\delta}(a))\subset f^{-1}(\mathbb{B}_{\varepsilon}(f(a))))$ (you can prove that this is true)

$\implies f(\mathbb{B}_{\delta}(a)))\subset \mathbb{B}_{\varepsilon}(f(a)))$

$||x-a||<\delta \implies ||f(x)-f(a)||<\varepsilon $

$\therefore f$ is continuous.

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